# Thread: Implicit differentiation of e^y?

1. ## Implicit differentiation of e^y?

Hello,

I was just trying to figure out one thing, and I was wondering...how do you find the implicit differentiation of e^y? Would I have to multiply it by dy/dx like you're supposed to do whenever you take the derivative of a variable that is y? I know the derivative of e^x is just e^x, so I'm assuming it would be the same for e^y, but I just wasn't sure if I was supposed to put dy/dx beside it....sorry if this seems like a silly question, I only just learned how to do implicit differentiation very recently.

Thanks!

p.s. If you were curious as to what the question was, it was e^y*cos(x) = 3 + sin(xy), and I was asked to find y'. So far I have:

e^y(dy/dx?) * (-sin(x)) = cos(y+x(dy/dx))

2. Originally Posted by dark-ryder341
Hello,

I was just trying to figure out one thing, and I was wondering...how do you find the implicit differentiation of e^y? Would I have to multiply it by dy/dx like you're supposed to do whenever you take the derivative of a variable that is y? I know the derivative of e^x is just e^x, so I'm assuming it would be the same for e^y, but I just wasn't sure if I was supposed to put dy/dx beside it....sorry if this seems like a silly question, I only just learned how to do implicit differentiation very recently.

Thanks!

p.s. If you were curious as to what the question was, it was e^y*cos(x) = 3 + sin(xy), and I was asked to find y'. So far I have:

e^y(dy/dx?) * (-sin(x)) = cos(y+x(dy/dx))

I can't understand what you did, but the implicit differentiation of your equation is:

$e^y\cos x\,dy-e^y\sin x\,dx=y\cos xy\,dx+x\cos xy\,dy\Longrightarrow \frac{dy}{dx}=\frac{e^y\sin x+y\cos xy}{e^y\cos x-x\cos xy}$

Tonio